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A solid brass sphere of volume $0.305\,{m^3}$ is dropped in the ocean, where water pressure is $2 \times {10^7}N/{m^2}$. The bulk modulus of water is $6.1 \times {10^{10}}\;N/{m^2}$. What is the change in volume of the sphere?

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Last updated date: 26th Feb 2024
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IVSAT 2024
Answer
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Hint: In order to solve this problem, we need to use the formula of bulk modulus and observe the situation that when an incompressible liquid is taken under consideration then what would be the value of the terms present over there. Doing this will solve our problem and we will get the right answer.

Complete step by step answer:
Bulk Modulus: Also referred to as incompressibility, the bulk modulus is a measure of the ability of a material to tolerate changes in volume when under compression on all sides. It is equal to the quotient of the applied pressure which is divided by the relative deformation.

We know that the change in volume $\Delta V$ is given as:
$\Delta V = \dfrac{{ - PV}}{K}$
Here, P is the pressure, V is the volume and K is the bulk modulus.
Substitute the known values in the given equation from the question given.
$\begin{array}{l}
\Delta V = \dfrac{{ - 2 \times {{10}^7}N/{m^2} \times 0.305\;{m^3}}}{{6 \times {{10}^{10}}\;N/{m^2}}}\\
               = - {10^4}\,{m^3}
\end{array}$

Note: In this problem you just need to recall the basic formula of bulk modulus and observe the change in volume when the liquid is incompressible. Such problems generally arise in competitive exams in which you have to use the formula of the term and have to give the answer according to that formula. Bulk Modulus is a constant that denes how resistant the material is to compression. It is defined as the ratio between the increase in pressure and the resulting decrease in the volume of the material. The density of the material is the degree of compactness of the material. This is known as the mass per unit volume.